Abstract

This paper aims to establish the Tikhonov regularization theory for set-valued
variational inequalities. For this purpose, we firstly prove a very general existence
result for set-valued variational inequalities, provided that the mapping involved
has the so-called variational inequality property and satisfies a rather weak coercivity
condition. The result on the Tikhonov regularization improves some known results
proved for single-valued mapping.

1. Introduction

This paper discusses the generalized variational inequality problem (in short, ) which is to find and such that
where is a nonempty closed convex set in and is a set-valued mapping with nonempty values. We use and to denote the problem (1.1) and its solution set, respectively.

Generalized variational inequality has been extensively studied in the literature; see [1–7] and the references therein.

The Tikhonov regularization method is an important method for the ill-posed variational inequalities; see Pages 307 and 1224 in [8]. To our best knowledge, the Tikhonov regularization method has been discussed only for the case where the mapping is single valued. This paper develops the Tikhonov regularization method for set-valued variational inequality .

As a preparation, we firstly give an existence result for . It is well known if is upper semicontinuous with nonempty compact convex values and the set is compact and convex, then has a solution. If is noncompact, one usually requires some kind of coerciveness conditions for the existence of solutions to . Thus many researchers have attempted to search coerciveness condition as weak as possible; see [2, 3, 8, 9] and the references therein. In particular, [9] proved the following result.

Theorem 1.1. If is single valued and continuous, and if the following condition is satisfied:(H) There exists such that for every , there is with satisfying ,then the variational inequality has a solution.

Example 3.1 in [9] shows that the condition (H) is strictly weaker than many known coerciveness conditions. So far, it is not known whether Theorem 1.1 could be extended to the situation where is a set-valued mapping. An affirmative answer is given in Corollary 3.9 of this paper, which says that if is upper semicontinuous with nonempty compact convex values, then the coerciveness condition (A), which reduces to the condition (H) when is single-valued, implies that has a solution. Actually, a more general existence result is verified in Theorem 3.8 which does not require that have any kind of continuity. Theorem 3.8 shows that if has the so-called variational inequality property, then the condition (A) implies that has a solution. If the mapping is either upper semicontinuous (with nonempty compact convex values), or quasimonotone and upper hemicontinuous (with nonempty compact convex values), then it has the variational inequality property. Thus Theorem 3.8 unifies many known existence results for .

The Tikhonov regularization method has been much discussed in the literature. In particular, assuming that is a box and is a single-valued mapping, [10] proved that is nonempty for any (here stands for the identity mapping), provided that is a continuous -function and is nonempty and bounded. This result was extended by [11] to the situation where is a closed convex set, the mapping is single valued, and a coercivity condition is used to replace the assumption of being bounded. It should be noted that the coercivity condition assumed in [11] does not necessarily imply that is bounded. [12] further improves the result of [11] by assuming a weaker coercivity condition. All the above results on the Tikhonov regularization assume that the mapping is single valued. The last part of this paper aims to establish the Tikhonov regularization theory for set-valued variational inequality (1.1). Theorem 4.1 improves the main result of [11, 12] by assuming a weaker coercivity condition and by allowing to be a set-valued mapping (without monotonicity).

General variational inequalities have been extensively discussed; see [13–16]. It should be interesting to discuss the Tikhonov regularization method for general variational inequality in a similar way.

2. Preliminaries

Unless stated otherwise, we assume that is a nonempty closed convex set and is a set-valued mapping with nonempty values. For , .

Definition 2.1. Let be a set-valued mapping. is said to be(i)monotone on if for each pair of points and for all and , ,(ii)maximal monotone on if, for any , for all and all implies ,(iii)quasimonotone on if for each pair of points and for all and , implies that ,(iv) is said to be upper semicontinuous at if for every open set containing , there is an open set containing such that for all ; if is upper semicontinuous at every , we say is upper semicontinuous on ,(v)upper hemicontinuous on if the restriction of to every line segment of is upper semicontinuous.

Lemma 2.2. Let be a nonempty convex subset of a Hausdorff topological vector space , and let be a set-valued mapping from into satisfying the following properties:(i) is a KKM mapping: for every finite subset of , , where denotes the convex hull;(ii) is closed in for every ;(iii) is compact in for some . Then .

3. Existence of Solutions and Coercivity Conditions

Definition 3.1. is said to have variational inequality property on if for every nonempty bounded closed convex subset of , has a solution.

Proposition 3.2. The following classes of mappings have the variational inequality property:(i)every upper semicontinuous set-valued mapping with nonempty compact convex values,(ii)every upper hemicontinuous quasimonotone set-valued mapping with nonempty compact convex values;(iii)if is a single-valued continuous mapping and is upper hemicontinuous and monotone with nonempty compact convex values, then has the variational inequality property.

Proof. (i) is well known in the literature. (ii) is verified in [18]. (iii) is a consequence of the Debrunner-Flor lemma [19] and [20, Theorem 41.1]. Indeed, let be a bounded closed convex subset of , and let
the normal cone of at . By [20, Theorem 41.1], is a maximal monotone mapping. By the Debrunner-Flor lemma [19], there is such that
Since is maximal monotone, . By the definition of , .

Proposition 3.3 below shows that Proposition 3.2(iii) can be extended to the case where is a set-valued mapping.

Proposition 3.3. If is upper semicontinuous with nonempty compact convex values and is monotone and upper hemicontinuous on with nonempty compact convex values, then has the variational inequality property on .

Proof. Let be a bounded closed convex subset of . Define by
Since is monotone, . Since is upper semicontinuous, is closed and hence compact in for every . Now we prove that is a KKM map. If not, there is and such that
This contradiction shows that is a KKM map, so is . By Lemma 2.2, there is . Fix any . Let . Then for small , and hence
Dividing on both sides, we have
Letting yields that , as is upper hemicontinuous. Since is arbitrary, :
Since and are compact and convex, the Sion minimax theorem implies the existence of and such that
Thus solves .

Before making further discussion, we need to state some coercivity conditions. The relationships of these coercivity conditions are well known in the literature; however, we provide the proof for completeness.

Consider the following coercivity conditions.(A)There exists such that for every , there is with satisfying .(B)There exists such that for every , there is satisfying .(C)There exists such that for every and every , there exists some such that .(D)There exists such that for every , there exists some such that .(E)There exists such that the set

is bounded, if nonempty.

Proposition 3.4. The following statements hold.(i)(C)(B) if is of convex values.(ii)(D)(B) if is quasimonotone.(iii)(E)(B)(A).

Proof. (i) By (C), for every , . Since is convex and is compact convex, the Kneser minimax theorem implies that
Since is upper semicontinuous and since is compact, there is such that
This verifies (B). (ii) The implication of (D) (B) is an immediate consequence of being quasimonotone. (iii) (E) (B). If , then for any , ; thus (B) holds. If , then by (E), there is such that . Thus (B) is still true. (B) (A). Let be such that (B) holds. Then , . By (B), there is such that . Obviously, . Thus (A) is verified with replaced by .

Remark 3.5. The coercivity condition (A) is actually (C’) in [3] where it is shown that if is quasimonotone and upper hemicontinuous with nonempty compact convex values, then (A) implies that has a solution. However, it seems unknown whether this assertion still holds if one replaces “quasimonotone and upper hemicontinuous” by “upper semicontinuous.” An affirmative answer is given by Corollary 3.9.

Remark 3.6. The coercivity condition (B) is the condition (C) in [3]. The condition (D) appears in [21]. The condition (E) appears essentially in Corollary 3.1 in [22]; see also Proposition 2.2.3 in [8]. If is single valued, then (E) reduces to Proposition 2.2.3(a) in [8].

Remark 3.7. Example 3.1 in [9] shows that (A) does not necessarily imply (B), even if is single-valued and continuous.

From the above discussion, (A) is the weakest coercivity condition among them. [9] proved if is single valued and continuous, then (A) implies that has a solution. Corollary 3.9 shows that this assertion still holds even if is a set-valued mapping.

Theorem 3.8. Let be a nonempty closed convex set, and let be a mapping with nonempty compact convex values. Suppose that (A) holds. If has the variational inequality property on , then has a solution.

Proof. Let . Since is bounded closed convex and has the variational inequality property, there is such that
If , then , and by assumption, there is with such that
Fix any . Since , there is such that . It follows that
Therefore, . Since is arbitrary, the conclusion is verified. (ii) If , then for any , there is such that
It follows that
Since is arbitrary, solves .

Corollary 3.9. Let be a nonempty closed convex set, and let be upper semicontinuous set-valued mapping with nonempty compact convex values. Suppose that (A) holds. Then has a solution.

Proof. By Proposition 3.2, has the variational inequality property. The conclusion follows immediately from Theorem 3.8.

Remark 3.10. Assuming that the mapping is single valued and continuous, Proposition 2.2.3 in [8] and Theorem 3.2 in [9] show that is nonempty if the conditions (E) and (A) hold, respectively. Therefore, Corollary 3.9 improves Proposition 2.2.3 in [8] and Theorem 3.2 in [9]: the mapping is set valued instead of single valued.

4. The Tikhonov Regularization

Theorem 4.1. Let be a nonempty closed convex set in , and let be upper semicontinuous with nonempty compact convex values. If assumption (A) holds, then for any ,(i) has a solution;(ii)the set is bounded.

Proof. (i) Let be as in assumption (A). We claim that for every , there is with satisfying
Granting this, we obtain that assumption (A) is satisfied with the mapping replaced by . Since is upper semicontinuous with nonempty compact convex values and is continuous and monotone, Theorem 3.8 and Proposition 3.3 imply that has a solution. Now we prove the claim. Since for every , there is with satisfying , we obtain
Let and . Then . If not, by assumption (A), there is with such that
Since and ,
Therefore , a contradiction.

Remark 4.2. Theorem 4.1 improves in two ways: the mapping is set valued instead of single valued; Theorem 1 in [12] used the coercivity condition (B) while our Theorem 4.1 uses the weaker coercivity condition (A).

Theorem 4.3. Let be a nonempty closed convex set in , and let be upper semicontinuous with nonempty compact convex values. Assume that there is a nonempty bounded closed convex set such that
Then for any , has a solution and the set is bounded.

Proof. Let be such that
Then for every ,
Since is bounded closed convex and since is upper semicontinuous, there is such that
Since , . Therefore, assumption (A) is satisfied. Then the conclusion follows from Theorem 4.1.

Remark 4.4. If is a single-valued mapping, and for some , then Theorem 4.3 reduces to Theorem 2.1 in [11].

For every , let , and we define

Theorem 4.5. Let be a nonempty closed convex set in , and let be upper semicontinuous with nonempty compact convex values. If assumption (A) holds, then

Proof. The set being nonempty follows from Theorem 4.1. Let . Then there are a sequence and such that . This means that for some ,
Since is upper semicontinuous with nonempty compact values, is compact. Without loss of generality, assume for some . Thus for every ,
It follows from (4.11) that .

Acknowledgments

This work is partially supported by the National Natural Science Foundation of China (no. 10701059) and by the Sichuan Youth Science and Technology Foundation.